An element \(^{226}_{88}X\) undergoes radioactive decay by emitting two alpha particles and a beta radiation....

Explanation

The question is about an element \(^{226}_{88}X\) undergoing radioactive decay by emitting two alpha particles and a beta radiation. The question asks for the correct nuclei that describe the product formed by the reaction. To understand this question, you need to know what radioactive decay is. Radioactive decay is when the nucleus of an unstable atom breaks down, releasing particles and energy. Alpha particles are made up of two protons and two neutrons and are represented by the symbol \(^{4}_{2}\alpha\). Beta particles are high-energy electrons that come from the decay of a neutron into a proton and an electron. They are represented by the symbol \(^{0}_{-1}\beta\). According to the question, the element \(^{226}_{88}X\) emits two alpha particles and a beta radiation during the decay process. We need to determine the correct nuclei that describe the product formed by the reaction. Option A, \(^{222}_{89}T\), is incorrect because the atomic number is 89, not 88. Option B, \(^{218}_{84}S\), is incorrect because the total mass number is 218, which is not equal to the initial mass number of 226. Option D, \(^{227}_{87}Q\), is incorrect because the atomic number is 87, not 85. Option C, \(^{218}_{85}R\), is the correct answer. The two alpha particles have a total mass number of 8 and a total atomic number of 4. The beta particle has a mass number of 0 and an atomic number of -1. When we subtract the total mass and atomic numbers of the emitted particles from the initial values, we get a final product of \(^{218}_{85}R\), which matches Option C. In summary, the correct nuclei that describe the product formed by the reaction are \(^{218}_{85}R\), which is Option C.